Holonomic Function Identities Christoph Koutschan Research Institute for Symbolic Computation Johannes Kepler Universit¨atLinz, Austria ACA 2008 Algebraic and Algorithmic Aspects of Differential and Integral Operators History 1990: D. Zeilberger’s “Holonomic Systems Approach” 1998: extensions and refinements by F. Chyzak 2008: Mathematica implementation by CK Notation • K: field of characteristic 0 • An = An(K): the n-th Weyl algebra 0 • Dx: differential operator w.r.t. x, i.e., Dx • f(x) = f (x) • Sn: shift operator w.r.t. n, i.e., Sn • f(n) = f(n + 1) • O: an Ore algebra • AnnO f: the ideal of annihilating operators of f in O, i.e., AnnO f = {P ∈ O | P • f = 0} Definition: Ore Algebra (1) Let F be a K-algebra (of “functions”), and let σ, δ ∈ EndK F with δ(fg) = σ(f)δ(g) + δ(f)g for all f, g ∈ F (skew Leibniz law). The endomorphism δ is called a σ-derivation. Let A be a K-subalgebra of F (e.g., A = K[x] or A = K(x)) and assume that σ, δ restrict to a σ-derivation on A. Define the skew polynomial ring O := A[∂; σ, δ]: • polynomials in ∂ with coefficients in A • usual addition • product that makes use of the commutation rule ∂a = σ(a)∂ + δ(a) for all a ∈ A Definition: Ore Algebra (2) We turn F into an O-module by defining an action of elements in O on a function f ∈ F: a • f := a · f, ∂ • f := δ(f). Remark: In special cases we define the action ∂ • f := σ(f). d Example 1: A = K[x], σ = 1, δ = dx . d Then K[x][Dx; 1, dx ] = K[x][Dx; 1,Dx] is the Weyl algebra A1. Example 2: A = K[n], σ(n) = n + 1, σ(c) = c for c ∈ K, δ = 0. Then K[n][Sn; Sn, 0] is a shift algebra. Example 3: K(n, x, y)[Sn; Sn, 0][Dx; 1,Dx][Dy; 1,Dy] Holonomic functions Definition: A function f(x1, . , xn) ∈ F is said to be holonomic if An/ AnnAn f is a holonomic module. Definition: r A sequence f(k1, . , kr) ∈ CN is holonomic if its multivariate generating function ∞ ∞ X X k1 kr F (x1, . , xr) = ··· f(k1, . , kr)x1 ··· xr . k1=0 kr=0 is a holonomic function. Properties of holonomic functions Closure properties: • sum • product • definite integration Elimination property: Given an ideal I in An s.t. An/I is holonomic; then for any choice of n + 1 among the 2n generators of An there exists a nonzero operator in I that depends only on these. In other words, we can eliminate n − 1 variables. Integration via elimination Given: AnnO f, the annihilator of a holonomic function f(x, y) in the Ore algebra O = K[x, y][Dx; 1,Dx][Dy; 1,Dy]. R b Task: Compute F (y) = a f(x, y) dx Since AnnO f is holonomic, there exists P ∈ AnnO f that does not contain x (“elimination property”): P (y, Dx,Dy) = Q(y, Dy) + DxR(y, Dx,Dy) R b Apply a ... dx to P • f = 0: h ix=b Q(y, Dy) • F (y) + R(y, Dx,Dy) • f(x, y) x=a Summation via elimination Given: AnnO f, the annihilator of a holonomic sequence f(k, n) in the Ore algebra O = K[k, n][Sk; Sk, 0][Sn; Sn, 0]. Pb Task: Compute F (n) = k=a f(k, n) By the elimination property there exists P ∈ AnnO f that does not contain k: P (n, Sk,Sn) = Q(n, Sn) + (Sk − 1)R(n, Sk,Sn) Sum over this equation: h ik=b Q(n, Sn) • F (n) + R(n, Sk,Sn) • f(k, n) k=a Example: Orthogonality of Hermite Polynomials (1) Compute the integral Z ∞ −x2 √ n e Hm(x)Hn(x) dx = δm,n π2 n! −∞ First we compute an annihilator of the integrand: ann = Annihilator[Exp[-x^2]*HermiteH[m,x]*HermiteH[n,x], {S[m], S[n], Der[x]}] {−2x + Dx + Sm + Sn, 2 Dx + 2m − 2n + (−2x)Dx + 2SnDx − 2, 2 Sn + 2n − 2xSn + 2, 3 2 Dx + 4mx − 4nx − 4x + (4n − 4m)Sn + −4x + 2m + 6n + 4 Dx} Example: Orthogonality of Hermite Polynomials (2) Next step is to compute a Gr¨obner basis w.r.t. lexicographical order in order to eliminate x: gb = OreGroebnerBasis[ ann, OreAlgebra[x, m, n, S[m], S[n], Der[x]], MonomialOrder -> Lexicographic] {2n − SmSn − SnDx + 2, 2m − SmDx − SmSn + 2, Dx − 2x + Sm + Sn } Example: Orthogonality of Hermite Polynomials (3) In the first operator, the part R = −SnDx, in the second x=∞ R = −SmDx. We have to check that [R • f]x=−∞ indeed vanishes: Limit[D[Exp[-x^2]*HermiteH[m,x]*HermiteH[n+1,x], x], x -> Infinity] 0 Hence we take the first two operators (which do not involve the integration variable x) and set R to 0: OrePolynomialSubstitute[Take[gb, 2], {Der[x] -> 0}] {2n − SmSn + 2, 2m − SmSn + 2} Example: Orthogonality of Hermite Polynomials (4) By computing a last Gr¨obner basis, we get the result: OreGroebnerBasis[%, OreAlgebra[m, n, S[m], S[n]]] {m − n, 2n − SmSn + 2} This proves that the right hand side can only be nonzero if m = n. By similar computations we obtain the recurrence 4n2 + 8n + 4 f(n) + (−4n − 6)f(n + 1) + f(n + 2) = 0 for the right hand side when we set m to n. √ √ Together with the initial values f(0) = π and f(1) = 2 π we have a full and simple description of the desired result. Definite integration with Takayama Given: AnnO f, the annihilator of a holonomic function f(x, y) (with natural boundaries) in the Ore algebra O = K[x, y][Dx; 1; Dx][Dy; 1,Dy]. R b Find: The annihilator of F (y) = a f(x, y)dx in the Ore algebra 0 O = K[y][Dy; 1,Dy] Find P ∈ AnnO f which can be written in the form P (x, y, Dx,Dy) = Q(y, Dy) + DxR(x, y, Dx,Dy) R b Apply a ... dx to P • f = 0: Z b Z b Q(y, Dy)f(x, y) dx + DxR(x, y, Dx,Dy)f(x, y) dx a a Hence Q(y, Dy)F (y) = 0 The operator Q can be computed with Takayama’s algorithm. Comparison Zeilberger: 1. eliminate x 2. reduce modulo DxO Takayama (variant due to Chyzak/Salvy): 1. reduce modulo DxO 2. eliminate x How to eliminate x? Problem: After reducing modulo DxO, no multiplication by x is allowed! Example: P + DxQ @ ·x @ mod DxO @ © @@R xP + (Dxx − 1)Q P mod DxO ·x ? ? xP − Q 6= xP Takayama’s algorithm 0 Eliminate x by computing a Groebner basis in the O -module w.r.t. α the basis x , α ∈ N: 2 2 2 x (1 + y) + xDx = x (1 + y) + Dxx − 1 ≡ x (1 + y) − 1 mod DxO −→ gives (−1, 0, 1 + y, 0,... ) x + DxDy + y ≡ x + y mod DxO −→ gives (y, 1, 0,... ) We have to include also multiples by xα to the generators of AnnO f: 2 2 2 x + xDxDy + xy = x + DxxDy − xDy + xy ≡ x − x(Dy + y) −→ gives (0,Dy + y, 1, 0,... ) Takayama’s algorithm Input: a set of generators {G1,...,Gm} for AnnO f 0 Output: AnnO F 1. set d = max1≤i≤m degx Gi 0 Sm α 2. set A = {G1,...,Gm} ∪ i=1{x Gi | 1 ≤ α ≤ degx Gi} 0 3. reduce all elements in A modulo DxO 4. compute a Groebner basis in the corresponding module eliminating x 5. if no (P, 0,..., 0) is found, increase d Since f is holonomic the algorithm is guaranteed to terminate. Example: Victor Moll’s irresistible integral (1) Compute a closed form for the integral Z ∞ 1 m+1 dx 0 (x4 + 2ax2 + 1) Again, we start by computing annihilating operators for the integrand: ann = Annihilator[1/(x^4 + 2*a*x^2 + 1)^(m+1), {Der[x], S[m], Der[a]}] 2 2 4 2 {2mx + 2x + x + 2ax + 1 Da, 4 2 x + 2ax + 1 Sm − 1, 3 3 4 2 4mx + 4x + 4ax + 4amx + x + 2ax + 1 Dx} Example: Victor Moll’s irresistible integral (2) Although the integral does not have natural boundaries (e.g. the integrand does not vanish for x = 0), Takayama’s algorithm gives the correct result here: Takayama[ann, {x}, Saturate -> True] {4m + (2a)Da + (−4m − 4)Sm + 3, 2 2 4m + 4a − 4 Da + (8ma + 12a)Da + 3} Example: Victor Moll’s irresistible integral (3) The second operator is an ordinary differential equation in a: de = ApplyOreOperator[%[[2]], Int[a]] (4m + 3)Int(a) + (8ma + 12a)Int0(a) + 4a2 − 4 Int00(a) This ODE can be automatically solved by using Mathematica’s DSolve command (also the initial values can be computed automatically), giving the final result m 1 −m− 1 m −m−2 2 − 2 − 4 3 2 (1 + i)i 2 a − 1 πΓ 2m + 2 Pm (a) Γ(m + 1) ∂-finite functions Definition: A function f(x1, . , xm) is called ∂-finite w.r.t. O = K(x1, . , xm)[∂1; σ1, δ1] ··· [∂m; σm, δm] if O/ AnnO f is a finite-dimensional K(x1, . , xm)-vector space. In other words, f is ∂-finite if its “derivatives” span a finite-dimensional K(x1, . , xm)-vector space. x+y Example: All derivatives (w.r.t. x and y) of sin x−y are of the form x + y x + y r (x, y) sin + r (x, y) cos , 1 x − y 2 x − y e.g., “ ” 32 3x4+12yx3−30y2x2−4y3x+9y4 sin x+y 3 2 x+y ( ) x−y Dx Dy • sin x−y = (x−y)9 5 4 3 2 4 5 “ x+y ” 16(6x −33yx +80y x −54y x+3y ) cos x−y − (x−y)10 Chyzak’s extension of Zeilberger’s fast algorithm Given: AnnO f, the annihilator of a ∂-finite function f(x, y) in the rational Ore algebra O = K(x, y)[Dx; 1; Dx][Dy; 1,Dy].